Movement of the field

**Apprentice123** · April 27th 2009, 11:43 AM

Find the movement of the field
$F(x,y) = (x^2-y)i + (x-y^2)j$

arrond the curve $\alpha$ which is the boundary of the region in 1° quedrant understood by coordinated axes and the circle $x^2+y^2=16$

My solution

$\int_0^{\frac{ \pi}{2}} (-64cos^2tsint +16sen^2t+16cos^2t-64sen^2tcost)dt = 8 \pi$

Correct ?

**Apprentice123** · May 1st 2009, 04:42 PM

I figured again and got:

$8 \pi \frac{128}{3}$

**TheEmptySet** · May 1st 2009, 04:53 PM

Originally Posted by Apprentice123

Find the movement of the field
$F(x,y) = (x^2-y)i + (x-y^2)j$

arrond the curve $\alpha$ which is the boundary of the region in 1° quedrant understood by coordinated axes and the circle $x^2+y^2=16$

My solution

$\int_0^{\frac{ \pi}{2}} (-64cos^2tsint +16sen^2t+16cos^2t-64sen^2tcost)dt = 8 \pi$

Correct ?

For this one you would want to use Greens' theorem:

$\iint_D 1-(-1)dA=2 \iint_D dA=2 \frac{1}{4}\pi (4)^2=8\pi$

**Apprentice123** · May 1st 2009, 05:16 PM

Thank you

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